# Solve the following problem : In the following probability distribution of a r.v.X. x 1 2 3 4 5 P (x) 120 320 a 2a 120 Find a and obtain the c.d.f. of X. - Mathematics and Statistics

Sum

Solve the following problem :

In the following probability distribution of a r.v.X.

 x 1 2 3 4 5 P (x) (1)/(20) (3)/(20) a 2a (1)/(20)

Find a and obtain the c.d.f. of X.

#### Solution

Since the given table represents a p.m.f. of r.v. X,

$\sum\limits_{x=1}^{5}\text{P}(x) = 1$

∴ P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 1

∴ (1)/(20) + (3)/(20) + "a" + 2"a" + (1)/(20) = 1

∴ 3a = 1 - (5)/(20)

∴ 3a = 1 - (1)/(4)

∴ 3a = (3)/(4)

∴ a = (1)/(4)
By definition of c.d.f.,
F(x) = P(X ≤ x)

F(1) = P(X ≤ 1) = P(1) = (1)/(20)

F(2) = P(X ≤ 2) = F(1) + P(2)

= (1)/(20) + (3)/(20) = (4)/(20)

F(3) = P(X ≤ 3)
= F(2) + P(3)

= (4)/(20) + "a" = (4)/(20) + (1)/(4) = (4)/(20) + (5)/(20) = (9)/(20)

F(4) = P(X ≤ 4)
= F(3) + P(4)

= (9)/(20) + 2"a" = (9)/(20) + (1)/(2) = (9)/(20) + (10)/(20) = (19)/(20)

F(5) = P(X ≤ 5)
= F(4) + P(5)

= (19)/(20) + (1)/(20) = 1
∴ c.d.f. of X is as follows:

 xi 1 2 3 4 5 F(xi) (1)/(20) (4)/(20) (9)/(20) (19)/(20) 1
Concept: Probability Distribution of a Continuous Random Variable
Is there an error in this question or solution?
Chapter 8: Probability Distributions - Part I [Page 155]

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